From the book "Calculus" by James Stewart,

The Closed Interval Method is used to find the absolute(global) maximum and minimum values of a continuous function on a close interval $[a,b]$.

1) Find the values of $f$ at the critical numbers of $f$ in $(a,b)$.

2) Find the values of $f$ at the endpoints of the interval.

3) The largest of the values from Steps 1 and 2 is the absolute maximum value; the smallest of these values is the absolute minimum value.

I have problem understanding step 1.

Fermat's Theorem states that if function $f$ has local maximum/minimum at point $x=a$, then $a$ is a critical number.

Also the converse is not true as the function $y=x^3$ has a critical point at $x=0$ but $x=0$ is not a local minimum/maximum.

Also, from my understanding, absolute minimum/maximum on an closed interval can either be a local min/max on the closed interval or at the end points.

Hence, we should find the local minimum/maximum and compare the values with the value of function at the endpoints.

But from the example of $y=x^3$, doing step 1) as suggested in the book can derive point that is a critical number but not a local minimum/maximum.

Perhaps, it is proven such points can never be the absolute maximum/minimum?

Edit: knowing such point $a$ is not a local minimum/maximum tells us that there exist $x$ such that $f(x) > a$ and $f(x) < a$ and the absolute maximum/minimum would either be another critical number or at the end points.


Step 1 is just part of the whole story. And you are right that Step 1 would give points that are not absolute min/max. That's exactly why you need Step 2 and Step 3.

In the case of $f(x)=x^3$ (say in $[-1,1]$), it is true that you will get $x=0$ as a critical number. However, Step 2 and Step 3 will rule it out.

[Added to answer the question in the comment:] Because global min/max must also be local min/max. If a critical point is not a local min/max, then it cannot be global min/max.

  • $\begingroup$ why is it that Step 2 and Step 3 will definitely rule out such points? Can you explain in more details? $\endgroup$ – Little Rookie Jan 4 '17 at 15:21
  • $\begingroup$ In step 1, you get $f(0)=0$. But in Step 2, you get $f(-1)=-1$ and $f(1)=1$. Now by step 3, what you get? $\endgroup$ – Jack Jan 4 '17 at 15:24
  • $\begingroup$ How is it that step 2 and step 3 will rule out such points in general? Not only in the case of $y=x^3$ $\endgroup$ – Little Rookie Jan 4 '17 at 15:28
  • $\begingroup$ It is because by definition so. In Step 3, one compares all the points one gets in Step 1 and Step 2 so that one can identify the absolute max and absolute min. The bad points you described would not "win" in the comparison because they cannot be the absolute max/min. $\endgroup$ – Jack Jan 4 '17 at 15:31
  • $\begingroup$ Intuitively, they cannot be the absolute maximum/minimum. But is there a formal proof that i can refer to? $\endgroup$ – Little Rookie Jan 4 '17 at 15:53

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.